# -*- coding: utf-8 -*-
# author yzs
# date 2019-01-14
#
# 牛的繁殖问题
# Description
# Cows in the FooLand city are interesting animals. One of their specialties is related to producing offsprings.
# A cow in FooLand produces its first calve (female calf) at the age of two years and proceeds to produce
# other calves (one female calf a year).
# Now the farmer Harold wants to know how many animals would he have at the end of N years,
# if we assume that none of the calves die, given that initially, he has only one female calf?
# explanation:At the end of 1 year, he will have only 1 cow, at the end of 2 years he will have 2 animals
# (one parent cow C1 and other baby calf B1 which is the offspring of cow C1).At the end of 3 years,
# he will have 3 animals (one parent cow C1 and 2 female calves B1 and B2, C1 is the parent of B1 and B2).
# At the end of 4 years, he will have 5 animals (one parent cow C1,
# 3 offsprings of C1 i.e. B1, B2, B3 and one offspring of B1).
# Input
# The first line contains a single integer T denoting the number of test cases.
# Each line of the test case contains a single integer N as described in the problem.
# Output
# For each test case print in new line the number of animals expected at the end of N years modulo 10^9 + 7.
# Sample Input 1 
# 2
# 2
# 4
# Sample Output 1
# 2
# 5


def fibonacci(n):
    if n == 1:
        count = 1
    elif n == 2:
        count = 2
    else:
        count = fibonacci(n - 1) + fibonacci(n - 2)
    return count


while True:
    try:
        t = int(input().strip())
        for i in range(t):
            n = int(input().strip())
            print(fibonacci(n))
    except EOFError:
        break
